Question

For non-coplanar vectors $A,B$ and $C,\left|\right[ABC\left]\right|=\left|A\right|\left|B\right|\left|C\right|$ holds if and only if

• A. $A\cdot B=B\cdot C=C\cdot A=0$
• B. $A\cdot B=0=B\cdot C$
• C. $A\cdot B=0=C\cdot A$
• D. $B\cdot C=0=C\cdot A$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $A\cdot B=B\cdot C=C\cdot A=0$
B $A\cdot B=0=B\cdot C$
C $A\cdot B=0=C\cdot A$
D $B\cdot C=0=C\cdot A$

Let $\theta$ be the angle between $A$ and $B$ and $\varphi$ the angle between $C$
and $A\times B$ i.e. the angle between $C$ and the perpendicular to the plane containing $A$ and $B$. Then
$\left|\right(A\times B)\cdot C|=|A\times B|\left|C\right|\cos \varphi$
$=\left|A\right|\left|B\right|\left|C\right|\sin \theta \cos \varphi$
so if the given relation holds, we have $\sin \theta \cos \varphi =1$ since
$\left|\sin \theta \right|\left|\cos \varphi \right|\le 1$ so we must have $\sin \theta =1,\cos \varphi =1$ i.e.
$\theta =\pi /2,\varphi =0$. The former implies that $A$ and $B$ are
perpendicular so that $A\cdot B=0$ On the hand, if $\varphi$ is zero, $C$ must be perpendicular to both $A$ and $B$
so that $B\cdot C=0=A\cdot C.$